Mixing
If $frac{∂v}{∂t}=fracpartial{∂y}left(frac{∂v}{∂y}+h'vright)$ with $v(0,y)=v_0(y)$ and $int...
$begingroup$
Suppose $v:[0,infty)timesmathbb Rtomathbb R$ is a solution of $$frac{partial v}{partial t}=fracpartial{partial y}left(frac{partial v}{partial y}+h'vright);,;;;v(0,y)=v_0(y)$$ for some $hin C^1(mathbb R)$ and $v_0in C^0(mathbb R)$ with $$int v_0(y):{rm d}y=1.tag1$$
Why are we able to conclude $$int v(t,y):{rm d}y=1tag2$$ for all $tge0$?
integration pde
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
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Suppose $v:[0,infty)timesmathbb Rtomathbb R$ is a solution of $$frac{partial v}{partial t}=fracpartial{partial y}left(frac{partial v}{partial y}+h'vright);,;;;v(0,y)=v_0(y)$$ for some $hin C^1(mathbb R)$ and $v_0in C^0(mathbb R)$ with $$int v_0(y):{rm d}y=1.tag1$$
Why are we able to conclude $$int v(t,y):{rm d}y=1tag2$$ for all $tge0$?
integration pde
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
$endgroup$
add a comment |
$begingroup$
Suppose $v:[0,infty)timesmathbb Rtomathbb R$ is a solution of $$frac{partial v}{partial t}=fracpartial{partial y}left(frac{partial v}{partial y}+h'vright);,;;;v(0,y)=v_0(y)$$ for some $hin C^1(mathbb R)$ and $v_0in C^0(mathbb R)$ with $$int v_0(y):{rm d}y=1.tag1$$
Why are we able to conclude $$int v(t,y):{rm d}y=1tag2$$ for all $tge0$?
integration pde
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
$endgroup$
Suppose $v:[0,infty)timesmathbb Rtomathbb R$ is a solution of $$frac{partial v}{partial t}=fracpartial{partial y}left(frac{partial v}{partial y}+h'vright);,;;;v(0,y)=v_0(y)$$ for some $hin C^1(mathbb R)$ and $v_0in C^0(mathbb R)$ with $$int v_0(y):{rm d}y=1.tag1$$
Why are we able to conclude $$int v(t,y):{rm d}y=1tag2$$ for all $tge0$?
integration pde
integration pde
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
asked Jan 26 at 18:24
0xbadf00d0xbadf00d
1,87341533
1,87341533
add a comment |
add a comment |
1 Answer
1
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oldest
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Compute using the fundamental theorem of calculus:
$$
frac{d}{dt} int v(t,y) dy = int partial_t v(t,y) dy = int partial_y( partial_y v(t,y) + h'(y) v(t,y))dy = lim_{z to infty} left[partial_y v(t,z) + h'(z) v(t,z) - partial_y v(t,-z) + h'(-z) v(t,-z) right] =0,
$$
assuming that $partial_y v$ and $v$ decay sufficiently rapidly at infinity. This can be justified, for instance, if the solution lives in an appropriate Sobolev space. Then since the time derivative vanishes, we have
$$
int v(t,y) dy = int v(0,y) dy = int v_0(y) dy = 1
$$
for all $t ge 0$.
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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$begingroup$
Compute using the fundamental theorem of calculus:
$$
frac{d}{dt} int v(t,y) dy = int partial_t v(t,y) dy = int partial_y( partial_y v(t,y) + h'(y) v(t,y))dy = lim_{z to infty} left[partial_y v(t,z) + h'(z) v(t,z) - partial_y v(t,-z) + h'(-z) v(t,-z) right] =0,
$$
assuming that $partial_y v$ and $v$ decay sufficiently rapidly at infinity. This can be justified, for instance, if the solution lives in an appropriate Sobolev space. Then since the time derivative vanishes, we have
$$
int v(t,y) dy = int v(0,y) dy = int v_0(y) dy = 1
$$
for all $t ge 0$.
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
$endgroup$
add a comment |
$begingroup$
Compute using the fundamental theorem of calculus:
$$
frac{d}{dt} int v(t,y) dy = int partial_t v(t,y) dy = int partial_y( partial_y v(t,y) + h'(y) v(t,y))dy = lim_{z to infty} left[partial_y v(t,z) + h'(z) v(t,z) - partial_y v(t,-z) + h'(-z) v(t,-z) right] =0,
$$
assuming that $partial_y v$ and $v$ decay sufficiently rapidly at infinity. This can be justified, for instance, if the solution lives in an appropriate Sobolev space. Then since the time derivative vanishes, we have
$$
int v(t,y) dy = int v(0,y) dy = int v_0(y) dy = 1
$$
for all $t ge 0$.
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
$endgroup$
add a comment |
$begingroup$
Compute using the fundamental theorem of calculus:
$$
frac{d}{dt} int v(t,y) dy = int partial_t v(t,y) dy = int partial_y( partial_y v(t,y) + h'(y) v(t,y))dy = lim_{z to infty} left[partial_y v(t,z) + h'(z) v(t,z) - partial_y v(t,-z) + h'(-z) v(t,-z) right] =0,
$$
assuming that $partial_y v$ and $v$ decay sufficiently rapidly at infinity. This can be justified, for instance, if the solution lives in an appropriate Sobolev space. Then since the time derivative vanishes, we have
$$
int v(t,y) dy = int v(0,y) dy = int v_0(y) dy = 1
$$
for all $t ge 0$.
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
$endgroup$
Compute using the fundamental theorem of calculus:
$$
frac{d}{dt} int v(t,y) dy = int partial_t v(t,y) dy = int partial_y( partial_y v(t,y) + h'(y) v(t,y))dy = lim_{z to infty} left[partial_y v(t,z) + h'(z) v(t,z) - partial_y v(t,-z) + h'(-z) v(t,-z) right] =0,
$$
assuming that $partial_y v$ and $v$ decay sufficiently rapidly at infinity. This can be justified, for instance, if the solution lives in an appropriate Sobolev space. Then since the time derivative vanishes, we have
$$
int v(t,y) dy = int v(0,y) dy = int v_0(y) dy = 1
$$
for all $t ge 0$.
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
answered Jan 26 at 21:46
GlitchGlitch
5,6681030
5,6681030
add a comment |
add a comment |
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